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Interesting note from the page: "Some well-known infinite integer sequences provably satisfy Benford's Law exactly including Fibonacci numbers, factorials, and

Interesting note from the page: "Some well-known infinite integer sequences provably satisfy Benford's Law exactly including Fibonacci numbers, factorials, and the powers of almost any number."

markc 9y ago
Had to try it! (Clojure) (frequencies (take 100 (map #(first (str (first %))) (iterate (fn [[x y]] [y (+' x y)]) [0 1])))) => {\0 1, \1 30, \2 18, \3 12, \4 9, \5 8, \6 6, \7 5, \8 7, \9 4} i.e. Yup, Fibonacci up to 100, 30% start with 1, 18% start with 2.
tgb 9y ago
What are the numbers whose powers don't satisfy the law? 0 of course and any power of 10 will have only leading 1's. Any product of just 2's and 5's, probably?
- madcaptenor 9y ago
  Powers of two satisfy Benford's law. In order for powers of an integer k to not satisfy Benford's law, k would have to be a rational power of 10; in fact it would have to be an integer power of 10 because if 10^(m/n) is rational then it's an integer. But a quick calculation makes it look like powers of two don't satisfy Benford's law, For example the first power of 2 which begins with 7 is 2^46, and the first power of 2 which begins with 9 is 2^53. This takes so long, roughly speaking, because 2^10 = 1024 is so close to a power of ten.
  - tgb 9y ago
    That surprised me since I saw the "generalization to digits beyond the first" in the article and it's clear that powers of 2 or 5 clearly don't have a Benford-style distribution of their last digits. However, that was a misunderstanding on my part, since that generalization says something about the first n digits, and any fixed n will eventually not include the last digit of the number.
    - madcaptenor 9y ago
      In fact powers of any integer don't have a Benford-style distribution of their last digits, since those digits will repeat periodicially. For example powers of 2 end in 2, 4, 8, 6, 2, 4, 8, 6, ...
  - mark-r 9y ago
    Yes they do. Python is a good resource for quick back-of-the-envelope calculations like this, I just tested the first digit of the powers of 2 from 0 to 9999. The percentages were 30.10, 17.61, 12.49, 9.70, 7.91, 6.70, 5.79, 5.12, and 4.58. These agree almost precisely with the Wikipedia numbers. Ditto for powers of 5.
    - madcaptenor 9y ago
      I agree. But for powers from 1 to 50 those same percentages are 30, 20, 10, 10, 10, 8, 2, 10, 0. In particular there is only one power of 2 in this range starting in 7 and none that ends in 9. This might make someone doing hand computation suspicious. (But really you can't tell much from such a small sample.)

Interesting note from the page: "Some well-known infinite integer sequences provably satisfy Benford's Law exactly including Fibonacci numbers, factorials, and the powers of almost any number."

markc 9y ago

Had to try it! (Clojure) (frequencies (take 100 (map #(first (str (first %))) (iterate (fn [[x y]] [y (+' x y)]) [0 1])))) => {\0 1, \1 30, \2 18, \3 12, \4 9, \5 8, \6 6, \7 5, \8 7, \9 4} i.e. Yup, Fibonacci up to 100, 30% start with 1, 18% start with 2.

tgb 9y ago

What are the numbers whose powers don't satisfy the law? 0 of course and any power of 10 will have only leading 1's. Any product of just 2's and 5's, probably?

madcaptenor 9y ago
Powers of two satisfy Benford's law. In order for powers of an integer k to not satisfy Benford's law, k would have to be a rational power of 10; in fact it would have to be an integer power of 10 because if 10^(m/n) is rational then it's an integer. But a quick calculation makes it look like powers of two don't satisfy Benford's law, For example the first power of 2 which begins with 7 is 2^46, and the first power of 2 which begins with 9 is 2^53. This takes so long, roughly speaking, because 2^10 = 1024 is so close to a power of ten.
- tgb 9y ago
  That surprised me since I saw the "generalization to digits beyond the first" in the article and it's clear that powers of 2 or 5 clearly don't have a Benford-style distribution of their last digits. However, that was a misunderstanding on my part, since that generalization says something about the first n digits, and any fixed n will eventually not include the last digit of the number.
  - madcaptenor 9y ago
    In fact powers of any integer don't have a Benford-style distribution of their last digits, since those digits will repeat periodicially. For example powers of 2 end in 2, 4, 8, 6, 2, 4, 8, 6, ...
- mark-r 9y ago
  Yes they do. Python is a good resource for quick back-of-the-envelope calculations like this, I just tested the first digit of the powers of 2 from 0 to 9999. The percentages were 30.10, 17.61, 12.49, 9.70, 7.91, 6.70, 5.79, 5.12, and 4.58. These agree almost precisely with the Wikipedia numbers. Ditto for powers of 5.
  - madcaptenor 9y ago
    I agree. But for powers from 1 to 50 those same percentages are 30, 20, 10, 10, 10, 8, 2, 10, 0. In particular there is only one power of 2 in this range starting in 7 and none that ends in 9. This might make someone doing hand computation suspicious. (But really you can't tell much from such a small sample.)